10 GOLFERS 6 ROUNDS

[JIMT23]

I have a group of 10 golfers heading south for 6 rounds. Could someone help with pairings that have least amount of same players playing together over the 6 rounds?
2 - threesomes, 1 foursome
Thanks in advance!

[Ian Wakeling]

I think the best solution to this problem is to have one player who always plays in the foursome.  Then it it possible to get some resonable balance for the other players.  For example:

  (4  1  7)  (3  2  6)  (8  9  5 10)
  (5  2  8)  (1  3  4)  (9  7  6 10)
  (6  3  9)  (2  1  5)  (7  8  4 10)
  (4  5  9)  (7  3  8)  (2  6  1 10)
  (5  6  7)  (8  1  9)  (3  4  2 10)
  (6  4  8)  (9  2  7)  (1  5  3 10)

In the above player 10 players with all the other players exactly twice.  The other 9 players all have exactly 4 games in a threesome and 2 games in a foursome. 

Hope that helps.